Basic Algebra - Factorising

Hi,

The numerator is 9x^2 - 4. According to the exam board the correct answer is b = 4. I think b = 2.
What do you think?

 

[math]ax^2- b^2= (x\sqrt{a}- b)(x\sqrt{a}+ b)[/math]. That COULD be (3x- 2)(3x+ 2) with a= 9 and b= 2.

In order that [math]9x^2- 4[/math] reduce to 3x- 2, the denominator must have a factor of 3x+ 2 so must be --- [math](3x+ 2)(4x- 1)= 12x^2 + 5x- 2[/math].

The fraction is [math]\frac{9x^2- 4}{12x^2+ 5x- 2}[/math] with a= 9, b= 4, c= 12, d= 5, and e= -2.

 

To factorize ax^2−b^2, we use the difference of squares formula:

ax^2−b^2=(x√a−b)(x√a+b).

Given that ax^2−b^2=9x2−4,

we identify a=9 and b=2.

This expression can be factored as (3x−2)(3x+2).

For 9x^2−4 to reduce to 3x−2, the denominator must contain a factor of 3x+2.

So, it should be of the form (3x+2)(4x−1)=12x^2+5x−2.

Thus, the fraction becomes 9x^2−4/12x^2+5x−2, where a=9, b=2, c=12, d=5, and e=−2.


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Thank you for your replies. It made sense.